Overflow in intermediate arithmetic
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
You are given a binary string s, and an integer k.
In one operation, you must choose exactly k different indices and flip each '0' to '1' and each '1' to '0'.
Return the minimum number of operations required to make all characters in the string equal to '1'. If it is not possible, return -1.
Example 1:
Input: s = "110", k = 1
Output: 1
Explanation:
'0' in s.k = 1, we can flip it directly in one operation.Example 2:
Input: s = "0101", k = 3
Output: 2
Explanation:
One optimal set of operations choosing k = 3 indices in each operation is:
[0, 1, 3]. s changes from "0101" to "1000".[1, 2, 3]. s changes from "1000" to "1111".Thus, the minimum number of operations is 2.
Example 3:
Input: s = "101", k = 2
Output: -1
Explanation:
Since k = 2 and s has only one '0', it is impossible to flip exactly k indices to make all '1'. Hence, the answer is -1.
Constraints:
1 <= s.length <= 105s[i] is either '0' or '1'.1 <= k <= s.lengthProblem summary: You are given a binary string s, and an integer k. In one operation, you must choose exactly k different indices and flip each '0' to '1' and each '1' to '0'. Return the minimum number of operations required to make all characters in the string equal to '1'. If it is not possible, return -1.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Math · Union-Find · Segment Tree
"110" 1
"0101" 3
"101" 2
Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3666: Minimum Operations to Equalize Binary String
class Solution {
public int minOperations(String s, int k) {
int n = s.length();
TreeSet<Integer>[] ts = new TreeSet[2];
Arrays.setAll(ts, i -> new TreeSet<>());
for (int i = 0; i <= n; i++) {
ts[i % 2].add(i);
}
int cnt0 = 0;
for (char c : s.toCharArray()) {
if (c == '0') {
cnt0++;
}
}
ts[cnt0 % 2].remove(cnt0);
Deque<Integer> q = new ArrayDeque<>();
q.offer(cnt0);
int ans = 0;
while (!q.isEmpty()) {
for (int size = q.size(); size > 0; --size) {
int cur = q.poll();
if (cur == 0) {
return ans;
}
int l = cur + k - 2 * Math.min(cur, k);
int r = cur + k - 2 * Math.max(k - n + cur, 0);
TreeSet<Integer> t = ts[l % 2];
Integer next = t.ceiling(l);
while (next != null && next <= r) {
q.offer(next);
t.remove(next);
next = t.ceiling(l);
}
}
ans++;
}
return -1;
}
}
// Accepted solution for LeetCode #3666: Minimum Operations to Equalize Binary String
func minOperations(s string, k int) int {
n := len(s)
ts := [2]*redblacktree.Tree{
redblacktree.NewWithIntComparator(),
redblacktree.NewWithIntComparator(),
}
for i := 0; i <= n; i++ {
ts[i%2].Put(i, struct{}{})
}
cnt0 := strings.Count(s, "0")
ts[cnt0%2].Remove(cnt0)
q := []int{cnt0}
ans := 0
for len(q) > 0 {
nq := []int{}
for _, cur := range q {
if cur == 0 {
return ans
}
l := cur + k - 2*min(cur, k)
r := cur + k - 2*max(k-n+cur, 0)
t := ts[l%2]
node, found := t.Ceiling(l)
for found && node.Key.(int) <= r {
val := node.Key.(int)
nq = append(nq, val)
t.Remove(val)
node, found = t.Ceiling(l)
}
}
q = nq
ans++
}
return -1
}
# Accepted solution for LeetCode #3666: Minimum Operations to Equalize Binary String
class Solution:
def minOperations(self, s: str, k: int) -> int:
n = len(s)
ts = [SortedSet() for _ in range(2)]
for i in range(n + 1):
ts[i % 2].add(i)
cnt0 = s.count('0')
ts[cnt0 % 2].remove(cnt0)
q = deque([cnt0])
ans = 0
while q:
for _ in range(len(q)):
cur = q.popleft()
if cur == 0:
return ans
l = cur + k - 2 * min(cur, k)
r = cur + k - 2 * max(k - n + cur, 0)
t = ts[l % 2]
j = t.bisect_left(l)
while j < len(t) and t[j] <= r:
q.append(t[j])
t.remove(t[j])
ans += 1
return -1
// Accepted solution for LeetCode #3666: Minimum Operations to Equalize Binary String
use std::collections::{BTreeSet, VecDeque};
impl Solution {
pub fn min_operations(s: String, k: i32) -> i32 {
let n: i32 = s.len() as i32;
let k: i32 = k;
let mut ts: [BTreeSet<i32>; 2] = [BTreeSet::new(), BTreeSet::new()];
for i in 0..=n {
ts[(i % 2) as usize].insert(i);
}
let cnt0: i32 = s.bytes().filter(|&c| c == b'0').count() as i32;
ts[(cnt0 % 2) as usize].remove(&cnt0);
let mut q: VecDeque<i32> = VecDeque::new();
q.push_back(cnt0);
let mut ans: i32 = 0;
while !q.is_empty() {
let size = q.len();
for _ in 0..size {
let cur = q.pop_front().unwrap();
if cur == 0 {
return ans;
}
let l = cur + k - 2 * cur.min(k);
let r = cur + k - 2 * (k - n + cur).max(0);
let parity = (l % 2) as usize;
let vals: Vec<i32> = ts[parity]
.range(l..=r)
.cloned()
.collect();
for v in vals {
q.push_back(v);
ts[parity].remove(&v);
}
}
ans += 1;
}
-1
}
}
// Accepted solution for LeetCode #3666: Minimum Operations to Equalize Binary String
import { AvlTree } from '@datastructures-js/binary-search-tree';
function minOperations(s: string, k: number): number {
const n: number = s.length;
const ts = [new AvlTree<number>(), new AvlTree<number>()];
for (let i = 0; i <= n; i++) {
ts[i % 2].insert(i);
}
let cnt0 = 0;
for (const c of s) {
if (c === '0') cnt0++;
}
ts[cnt0 % 2].remove(cnt0);
let q: number[] = [cnt0];
let ans = 0;
while (q.length > 0) {
const nq: number[] = [];
for (const cur of q) {
if (cur === 0) {
return ans;
}
const l = cur + k - 2 * Math.min(cur, k);
const r = cur + k - 2 * Math.max(k - n + cur, 0);
const t = ts[l % 2];
let node = t.upperBound(l, true);
while (node && node.getValue() <= r) {
const val = node.getValue();
nq.push(val);
t.remove(val);
node = t.upperBound(l, false);
}
}
q = nq;
ans++;
}
return -1;
}
Use this to step through a reusable interview workflow for this problem.
Track components with a list or adjacency matrix. Each union operation may need to update all n elements’ component labels, giving O(n) per union. For n union operations total: O(n²). Find is O(1) with direct lookup, but union dominates.
With path compression and union by rank, each find/union operation takes O(α(n)) amortized time, where α is the inverse Ackermann function — effectively constant. Space is O(n) for the parent and rank arrays. For m operations on n elements: O(m × α(n)) total.
Review these before coding to avoid predictable interview regressions.
Wrong move: Temporary multiplications exceed integer bounds.
Usually fails on: Large inputs wrap around unexpectedly.
Fix: Use wider types, modular arithmetic, or rearranged operations.